Question

19) Explain the variations in the following properties of group is elements. a) Atomic and ionic radii b) electronegativity c) Ionisation enthalpy D​

Answer 1

L=2m,

L=2m,d=3mm,A=

L=2m,d=3mm,A= 4

L=2m,d=3mm,A= 49π

L=2m,d=3mm,A= 49π

L=2m,d=3mm,A= 49π ×10

L=2m,d=3mm,A= 49π ×10 −6

L=2m,d=3mm,A= 49π ×10 −6 m

L=2m,d=3mm,A= 49π ×10 −6 m 2

L=2m,d=3mm,A= 49π ×10 −6 m 2

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL=

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 4

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm

hope it helps